POJ 1459：Power Network（最大流）

http://poj.org/problem?id=1459

题意：有np个发电站，nc个消费者，m条边，边有容量限制，发电站有产能上限，消费者有需求上限问最大流量。

思路：S和发电站相连，边权是产能上限，消费者和T相连，边权是需求上限，边的话就按题意加就好了。难点更觉得在于输入。。加个空格。。边数组要*2，因为有反向边。

  1 #include <cstdio>
  2 #include <algorithm>
  3 #include <iostream>
  4 #include <cstring>
  5 #include <string>
  6 #include <cmath>
  7 #include <queue>
  8 #include <vector>
  9 #include <map>
 10 #include <set>
 11 using namespace std;
 12 #define INF 0x3f3f3f3f
 13 #define N 110
 14 typedef long long LL;
 15 struct Edge {
 16     int v, cap, nxt;
 17     Edge () {}
 18     Edge (int v, int cap, int nxt) : v(v), cap(cap), nxt(nxt) {}
 19 }edge[N*N*2];
 20 int tot, cur[N], dis[N], pre[N], head[N], gap[N], S, T;
 21
 22 void Add(int u, int v, int cap) {
 23     edge[tot] = Edge(v, cap, head[u]); head[u] = tot++;
 24     edge[tot] = Edge(u, 0, head[v]); head[v] = tot++;
 25 }
 26
 27 void BFS() {
 28     queue<int> que;
 29     while(!que.empty()) que.pop();
 30     memset(dis, -1, sizeof(dis));
 31     memset(gap, 0, sizeof(gap));
 32     dis[T] = 0; gap[0]++; que.push(T);
 33     while(!que.empty()) {
 34         int u = que.front(); que.pop();
 35         for(int i = head[u]; ~i; i = edge[i].nxt) {
 36             int v = edge[i].v;
 37             if(dis[v] == -1) continue;
 38             dis[v] = dis[u] + 1;
 39             gap[dis[v]]++;
 40             que.push(v);
 41         }
 42     }
 43 }
 44
 45 int ISAP(int n) {
 46     memcpy(cur, head, sizeof(cur));
 47     BFS();
 48     int u = pre[S] = S, ans = 0, i;
 49     while(dis[S] < n) {
 50         if(u == T) {
 51             int flow = INF, index;
 52             for(int i = S; i != T; i = edge[cur[i]].v) {
 53                 Edge& e = edge[cur[i]];
 54                 if(e.cap < flow) {
 55                     flow = e.cap; index = i;
 56                 }
 57             }
 58             for(int i = S; i != T; i = edge[cur[i]].v) {
 59                 edge[cur[i]].cap -= flow;
 60                 edge[cur[i]^1].cap += flow;
 61             }
 62             ans += flow; u = index;
 63         }
 64         for(i = cur[u]; ~i; i = edge[i].nxt)
 65             if(dis[edge[i].v] == dis[u] - 1 && edge[i].cap > 0)
 66                 break;
 67         if(~i) {
 68             cur[u] = i;
 69             pre[edge[i].v] = u;
 70             u = edge[i].v;
 71         } else {
 72             int md = n;
 73             if(--gap[dis[u]] == 0) break;
 74             for(int i = head[u]; ~i; i = edge[i].nxt) {
 75                 if(md > dis[edge[i].v] && edge[i].cap > 0) {
 76                     md = dis[edge[i].v]; cur[u] = i;
 77                 }
 78             }
 79             ++gap[dis[u] = md + 1];
 80             u = pre[u];
 81         }
 82     }
 83     return ans;
 84 }
 85
 86 int main() {
 87     int n, nc, np, m;
 88     while(~scanf("%d%d%d%d", &n, &np, &nc, &m)) {
 89         tot = 0; memset(head, -1, sizeof(head));
 90         S = n, T = n + 1;
 91         int a, b, c;
 92         for(int i = 1; i <= m; i++) {
 93             scanf(" (%d,%d)%d", &a, &b, &c);
 94             Add(a, b, c);
 95         }
 96         for(int i = 1; i <= np; i++) {
 97             scanf(" (%d)%d", &a, &b);
 98             Add(S, a, b);
 99         }
100         for(int i = 1; i <= nc; i++) {
101             scanf(" (%d)%d", &a, &b);
102             Add(a, T, b);
103         }
104         printf("%d\n", ISAP(T+1));
105     }
106     return 0;
107 }